$\begin{aligned} & \text {Let } S=\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots \\ & =\frac{1}{3}\left(\frac{3}{4}-\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots\right)\end{aligned}$


Now, we know that
$$
(1+x)^{-n}=1-n x+\frac{n(n+1)}{2 !} x^2 \ldots \ldots
$$

$$
\begin{aligned}
& \text { Put } n=\frac{3}{2} \text { and } n=\frac{1}{2} \\
& \begin{aligned}
\Rightarrow \quad \frac{3}{2} \cdot \frac{1}{2} & -\frac{\frac{3}{2} \cdot \frac{5}{2}}{2 !}\left(\frac{1}{2}\right)^2+\ldots \\
= & 1-\left(1+\frac{1}{2}\right)^{-3 / 2}=1-\left(\frac{3}{2}\right)^{-3 / 2} \\
& =1-\left(\frac{2}{3}\right)^{3 / 2}=1-\frac{2 \sqrt{2}}{3 \sqrt{3}}=\frac{3 \sqrt{3}-2 \sqrt{2}}{3 \sqrt{3}}
\end{aligned}
\end{aligned}
$$
put in Eq. (i), we get
$$
S=\frac{1}{3}\left(\frac{3 \sqrt{3}-2 \sqrt{2}}{3 \sqrt{3}}\right) \Rightarrow S=\frac{3 \sqrt{3}-2 \sqrt{2}}{9 \sqrt{3}}
$$